ural 1246. Tethered Dog

1246. Tethered Dog

Time limit: 1.0 second
Memory limit: 64 MB
A dog is tethered to a pole with a rope. The pole is located inside a fenced polygon (not necessarily convex) with nonzero area. The fence has no self-crosses. The Olympian runs along the fence bypassing the vertices of the polygon in a certain order which is not broken during the jog. A dog pursues him inside the fenced territory and barks. Your program is to determine how (clockwise or counter-clockwise) the rope will wind after several rounds of the Olympian's jog.

Input

The first input line contains a number N that is the number of the polygon vertices. It’s known that 3 ≤ N ≤ 200000. The next N lines consist of the vertices plane coordinates, given in an order of Olympian’s dog. The coordinates are a pair of integers separated with a space. The absolute value of each coordinate doesn’t exceed 50000.

Output

You are to output "cw", if the rope is winded in a clockwise order and "ccw" otherwise.

Sample

inputoutput
4
0 0
0 1
1 1
1 0
 
cw
Problem Author: Evgeny Kobzev
Problem Source: Ural State University Personal Programming Contest, March 1, 2003 
Difficulty: 364 
 
题意：问一个多边形的给出顺序是不是顺时针。
分析：叉积判断方向。用最左或最优的点作为基准点。
  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <deque>
  6 #include <vector>
  7 #include <queue>
  8 #include <iostream>
  9 #include <algorithm>
 10 #include <map>
 11 #include <set>
 12 #include <ctime>
 13 #include <iomanip>
 14 using namespace std;
 15 typedef long long LL;
 16 typedef double DB;
 17 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 18 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 19 #define Rep(i, t) for(int i = (0); i < (t); i++)
 20 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 21 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 22 #define MIT (2147483647)
 23 #define INF (1000000001)
 24 #define MLL (1000000000000000001LL)
 25 #define sz(x) ((int) (x).size())
 26 #define clr(x, y) memset(x, y, sizeof(x))
 27 #define puf push_front
 28 #define pub push_back
 29 #define pof pop_front
 30 #define pob pop_back
 31 #define ft first
 32 #define sd second
 33 #define mk make_pair
 34 inline void SetIO(string Name)
 35 {
 36     string Input = Name+".in",
 37     Output = Name+".out";
 38     freopen(Input.c_str(), "r", stdin),
 39     freopen(Output.c_str(), "w", stdout);
 40 }
 41 
 42 inline int Getint()
 43 {
 44     int Ret = 0;
 45     char Ch = ' ';
 46     bool Flag = 0;
 47     while(!(Ch >= '0' && Ch <= '9'))
 48     {
 49         if(Ch == '-') Flag ^= 1;
 50         Ch = getchar();
 51     }
 52     while(Ch >= '0' && Ch <= '9')
 53     {
 54         Ret = Ret * 10 + Ch - '0';
 55         Ch = getchar();
 56     }
 57     return Flag ? -Ret : Ret;
 58 }
 59 
 60 const int N = 200010;
 61 struct Point
 62 {
 63     DB x, y;
 64     inline void Read()
 65     {
 66         scanf("%lf%lf", &x, &y);
 67     }
 68 } Arr[N];
 69 int n;
 70 
 71 inline void Input() 
 72 {
 73     scanf("%d", &n);
 74     For(i, 1, n) Arr[i].Read();
 75 }
 76 
 77 inline DB Det(const Point &A, const Point &O, const Point &B)
 78 {
 79     DB X1 = A.x - O.x, X2 = B.x - O.x;
 80     DB Y1 = A.y - O.y, Y2 = B.y - O.y;
 81     return X1 * Y2 - X2 * Y1;
 82 }
 83 
 84 inline void Solve()
 85 {
 86     int p = 1;
 87     For(i, 2, n)
 88         if(Arr[i].x > Arr[p].x) p = i;
 89     Arr[0] = Arr[n], Arr[n + 1] = Arr[1];
 90     if(Det(Arr[p - 1], Arr[p], Arr[p + 1]) >= 0.0) puts("cw");
 91     else puts("ccw");
 92 }
 93 
 94 int main()
 95 {
 96     #ifndef ONLINE_JUDGE 
 97     SetIO("E");
 98     #endif 
 99     Input();
100     Solve();
101     return 0;
102 }
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posted @ 2015-11-14 16:08  yanzx6  阅读(199)  评论(0)  编辑  收藏  举报